Class 11th Chemistry - Some Basic Concept of Chemistry Case Study Questions and Answers 2022 - 2023 | QB365

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Class 11th Chemistry - Some Basic Concept of Chemistry Case Study Questions and Answers 2022 - 2023

Published on: 09/09/2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Chemistry Subject - Some Basic Concept of Chemistry, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Questions:

Case Study Questions

3 x 4 = 12

Chemistry play an important role in human needs for food, health care products and improving life. Cis platin and taxol are used in chemotherapy, AZT (Azidothymidine) is used for AIDS. SI units are international units of measurement. Matter is classified into elements, compounds and mixtures, which can be homogeneous as well as heterogeneous. A mixture can be separated by physical methods, compounds can be separated by chemical methods only. Atomic mass is average of masses of isotopes depending upon their natural abundance. Empirical formula is calculated with the help of percentage composition of elements in a compound and molecular mass helps to calculate molecular formula. A chemical equation must be balanced so as to follow laws of chemical combination.
(a) Express 2.54 mm into S.I units.
(b) Out of milk, diamond, air, petrol which is pure substance?
(c) Balance the equation: NO₂ + H₂ O \(\rightarrow\) HNO₃ + NO
(d) What is percentage of Na in Na₂CO₃ ? (Na = 23u, C = 12, O = 16u)
(e) \({ }_{17}^{35} \mathrm{Cl} \text { and }{ }_{17}^{37} \mathrm{Cl}\) are in ratio of 3 : 1 in nature. What is atomic mass of CI?
(f) What is empirical formula of C₆H₁₂O₆?
(g) Chlorophyll contains 2.68% magnesium atoms. Calculate mass of magnesium atoms in 2 g of chlorophyll.

In the table given below to illustrate precision and accuracy. Study the table and answer the questions based on the table and related studied concepts.
Data to Illustrate Precision and Accuracy

Measurement in g	I	II	III	Average
student A	0.521 g	0.515 g	0.509 g	0.515 g
student B	0.516 g	0.515 g	0.514 g	0.515 g
student C	0.521 g	0.520 g	0.520 g	0.520 g

(a) What is meant by precision?
(b) What is accuracy?
(c) If actual mass of a piece of metal is 0.520 g, data for which student is neither precise nor accurate.
(d) Which student data is precise but not accurate?
(e) The data of which student is both precise and accurate?
(f) How many significant figures are in 0.520?
(g) What is scientific notation for 0.520?

Stoichiometry is a section of chemistry that involves calculation based on chemical equations. Chemical equations are governed by laws of chemical combination. Mass of reactants is equal to mass of products. Compound obtained from different methods contain the same elements in the fixed ratio by mass. Mole is a counting unit, equal to 6.022 x 10²³ particles.
One mole is also equal to molar mass expressed in grams. One mole of every gas at STP has volume equal to 22.4 L. The reacting species which are consumed in the reaction completely is called limiting reagent which decides amount of products formed. Concentration of solution is expressed in terms of molarity, molality and mole fraction.
(a) Calculate number of moles of NH₃ formed by reaction of 2 moles of N₂ and 2 moles of H₂.
\(\mathbf{N}_{2}(g)+3 \mathbf{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)
(b) Calculate number of electrons in 18 g of H₂ O. [Atomic number of H = 1 ,O = 8]
(c) Calculate the molality of 1 M NaCI solution having density 1.10 g cm^-3. (Molar mass = 58.5 g mol^-1)
(d) Define mole fraction.
(e) In aqueous solution of glucose X_glucose = 0.1, what is X_H2O = ?

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Answers Key:

Case Study Questions

3 x 4 = 12

(a) 2.54 x 10^{- 3} m.
(b) Diamond.
(c) 3NO₂ + H₂O \(\rightarrow\) 2HNO₃ + NO
(d) \(\% \text { of } \mathrm{Na}=\frac{\text { Total mass of } \mathrm{Na}}{\text { Molar mass }} \times 100\)
\(=\frac{46}{106} \times 100=43.39 \%\)
(e) \(\frac{3 \times 35+1 \times 37}{4}=35.5\)
(f) CH₂O
(g) Mass of magnesium atoms \(=2 \times \frac{2.68}{100}\)
\(=\frac{5.36}{100}=0.0536 \mathrm{~g}\)

(a) It refers to the closeness of the set of values obtained from identical measurements.
(b) It refers to the closeness of a single measurement to its true value.
(c) 'A' because the individual values differ widely and average value is not accurate.
(d) 'B'. The values deviate a little from each other but average is not equal to true value.
(e) 'C' because the value are close to each other as well as average is same as true value.
(f) 3
(g) 5.20 x 10^{- 3}

(a) 1 mole of N₂ needs 3 moles of H₂ ,
Therefore, 2 moles of N₂ needs 6 moles of H₂.
But we have only 2 moles of H₂ , so H₂is limiting reagent.
3 moles of H₂ gives 2 moles of NH₂.
2 moles of H₂gives \(\frac{2}{3} \times 2=1.33\) moles of NH₃ .
(b) 18 g of water (1 mole) containing 10 x 6.022 x 10²³electrons = 6.022 x 10²⁴ electrons.
[ \(\because\)1 molecule of H₂O = 2 + 8 = 10 e)
(c) 'M is molarity
'd is density of solution
'm' is molality
\(m=\frac{M \times 1000}{1000 \times d-M \times \text { Molar mass }}\)
\(=\frac{1 \times 1000}{1000 \times 1.10-1 \times 58.5}\)
\(=\frac{1000}{1100-58.5}=\frac{1000}{1041.5}=m=0.96 \mathrm{~mol} / \mathrm{kg} .\)
(d) It is ratio of number of moles of solute (component) to the total number of moles of solute and solvent (all components).
(e) X_glucose + x_H2O = 1
\(\Rightarrow\) X_H2O = 1 - 0.1 = 0.9